222a,0?y?a,0?z?a. 333121222222证明 由x?y?(a?x?y)?a有x?(y?a)x?(y?ay?a)?0. 其判别
24122222式??(y?a)?4(y?ay?a)?0(因x?R). 从而, 3y?2ay?0即0?y?a.
3422同理可证0?x?a,0?z?a.
330?x?3．设a,b,c表示一个三角形三边的长, 求证:
a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?3abc.
证明不失一般性, 设a?b?c, 令a?c?m,b?c?n, 则m?n?0. 有
3abc?a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?a(a?b)(a?c)?b(b?c)(b?a)?c(c?a)(c?b)?(c?m)(m?n)m?(c?n)n(n?m)?cmn?(m?n)[c(m?n)?(m2?n2)]?cmn?0.
?a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?3abc.
4．设x,y?R, 且x?y?1.求证: x2?2xy?y2?证明 设x?y??, 则由题设可知,
222222.
??1, 并可设x??cos?,x??sin?.于是
x2?2xy?y2??2(cos2??2cos?sin??sin2?)??2(cos2??sin2?)??22sin(??).
4?x2?2xy?y2?2.
5．已知a?1,b?1, 求证
?a?b?1.
1?ab证明 欲证
a?ba?b2?1成立, 只需()?1, 即证(a?b)2?(1?ab)2.
1?ab1?ab222222则只需(1?ab)?(a?b)?0, 也就是1?ab?a?b?0, 即证
(1?a2)(1?b2)?0. 而a?1,b?1, 所以(1?a2)(1?b2)?0成立. 命题得证.
6．若
?ai?1(ai?0), 求?(ai?i?1i?1nn11)?(n?)n. ain证明 a1?11111?a1?2?2?...?2?(n2?1)n2?12n2n2?1, a1na1na1na1na1n2项a2?11111?a2?2?2?...?2?(n2?1)n2?12n2n2?1, …… …… a2na2na2na2na2n2项an?11111?an?2?2?...?2?(n2?1)n2?12n2n2?1. annannannannann2项以上诸式, 当且仅当ai?1(i?1,2,...,n)是等号成立. 诸式两端相乘得n(a1?1111)(a2?)...(an?)?(n2?1)nnn2?12n3.
n2?1a1a2ann(a1a2...an)nn11n2?1n3?n)?n. 由已知?ai?1可得a1a2...an?,(naa...ai?112nn11n11112nn2?1(a?)?(n?). 即(a1?)(a2?)...(an?)?(n?1)n,?i2n3n?n3ana1a2annni?1i等号当且仅当a1?a2?...?an?851时成立. n27．证明: 函数f(x)?x?x?x?x?1?0. 证明 (1) 当x?(??,0)时, 显然f(x)?0;
(2) 当x?(0,1)时, f(x)?x?x(1?x)?(1?x)?0; (3) x?[1,??)时, f(x)?x(x?1)?x(x?1)?1?0. 综合(1), (2), (3)可知, 可知f(x)恒正. 8．证明 若ai?1(i?1,2,...,n),则2证明 用数学归纳法证明如下: 当n?1时, 命题显然成立;
假设命题对n成立, 我们来证明它对n?1也成立, 注意到ai?1(i?1,2,...,n).
n?182353(a1a2...an?1)?(1?a1)(1?a2)...(1?an).
?(1?a)?(1?aii?1n?1n?1)?2n?1(?ai?1)?2i?1nn?1(?ai??ai?an?1?1)i?1i?1n?1n?2[?ai?1?(?ai?an?1)]?2[(?ai?1)?(?ai?1??ai?1??ai?an?1)]n?1n?1i?1i?1i?1i?1i?1i?1n?1nn?1n?1n?1n?2(?ai?1)?2ni?1nn?1i?1n?1n?1(?ai?1??ai?an?1)i?1ni?1nn?1i?1n?1nn?1n?1?2(?ai?1)?2[?ai(an?1?1)?(an?1?1)]?2(?ai?1)?2(an?1)[?ai?1)n?1i?1i?1?2(?ai?1).
ni?1n?1故命题对n?1成立.
2n(1?a1?a2?...?an). 9．设ai?1(i?1,2,...,n), 求证?(1?ai)?n?1i?1n证明
nnai?1ai?1ai?1nn(1?a)?2(1?)?2(1?)?2(1?)????i22n?1i?1i?1i?1i?1nnn2n1(1??ai). ?2?[(n?1??(ai?1)]n?1n?1i?1i?1nnn10．设x?y?z?0,求证: 6(x?y?z)?(x?y?z).
证明 显然x?y?z?0是平凡情形. 假定x,y,z不全为零, 不妨设x?0,y?0.由
3332223z??(x?y), 得x3?y3?z3?3xyz. 记 I?6(x?y?z)?5x4yz?33332222xyxy2?16??z222?xyxy2???z??22?216???(2z2?2xy)3.
3??????再注意到x?y?(x?y)?2xy?z?2xy, 因而2z?2xy?x?y?z, 这就是所要证的不等式.
11．已知a,b为小于1的正数, 求证:
22222222a2?b2?(1?a)2?b2?a2?(1?b)2?(1?a)2?(1?b)2?22.
证明 设z1?a?bi,z2?(1?a)?bi,z3?a?(1?b)i,z4?(1?a)?(1?b)i,则
z1?a2?b2, z2?(1?a)2?b2,z3?(1?a)2?b2,
z4?a2?(1?b)2?(1?a)2?(1?b)2. z1?z2?z3?z4?z1?z2?z3?z4?2?2i?22.
?a2?b2?(1?a)2?b2?a2?(1?b)2?(1?a)2?(1?b)2?22.
,,R?,求证: a?b?c?abc?abc?abc,其中n?N,p,q,r, 12．设abc且p?q?r?n.
?nnnpqrqrprpqpan?qbn?rcn证明abc?na...a?b...b?c...c?,
nrpqpqrnnnnnnqan?rbn?pcnrpqran?pbn?qcn,abc?. 同理abc?nnqrp三式相加, 即得a?b?c?abc?abc?abc.
33322213．设a,b,c?R,求证: a?b?c?ab?bc?ca.
?nnnpqrqrprpq证明 该不等式关于a,b,c对称, 不妨设a?b?c,则由左式?右式
?a2(a?b)?b2(b?c)?c2(c?a)?a2(a?b)?b2(b?c)?c2(c?b?b?a)?(a2?c2)(a?b)?(b2?c2)(b?c)?0.
故a?b?c?ab?bc?ca. 14．已知a?b?0,求证3a?3b?证 欲证3a?3b?33332223a?b.
3a?b,由于a?b?0,所以3a?3b?0, a?b?0, 只要证
a?33a2b?33ab2?b?a?b, 即要证3ab2?3a2b, 由于3ab?0,只要证3b?3a,
由于a?b?0, 此不等式显然成立.
15．若p?R且p?2,不等式?log2x??plog2x?1?2log2x?p恒成立, 求实数x的取值范围.
解 令log2x?a,将不等式转化为: (a?1)p?a?2a?1?0,令
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